The mass of potassium dichromate crystals required to oxidise $750 {\mathrm{cm}}^3$ of 0.6 M Mohr's salt solution is (Given molar masses; potassium dichromate $=294\mathrm{g}{\mathrm{mol}}^{-1},\text{ Mohr's salt }=392\mathrm{g}{\mathrm{mol}}^{-1}\text{ ) }$

The reaction is ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+6{\mathrm{Fe}}^{2+}+14{\mathrm{H}}^{+}\to 2{\mathrm{Cr}}^{3+}+6{\mathrm{Fe}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$. From this reaction, it follows that 

$1 \mathrm{mol}\text{ of }{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}\equiv 6\mathrm{mol}\text{ of }{\mathrm{Fe}}^{2+}$  i.e. $1 \mathrm{mol}\text{ of }{\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\equiv 6 \mathrm{mol}\text{ of Mohr's salt }$

Amount of Mohr's salt in the given solution is  $n=VM=\left(0.750\mathrm{L}\right)(0.6\mathrm{molL}-1)=0.45\mathrm{mol}$

Amount of ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ required for the oxidation of ${\mathrm{Fe}}^{2+}$ ions will be $n_1=n/6=\left(0.45\mathrm{mol}\right)/6$

Mass of required ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ would be $m=n_1{M}_{\mathrm{m}}=\left(\frac{0.45}{6}\mathrm{mol}\right)\left(294\mathrm{g}{\mathrm{mol}}^{-1}\right)=22.05\mathrm{g}$